To complete the table for the quadratic equation \( y = 3x^2 - 6x + 9 \), we first calculate the values of \( y \) for \( x = -3, -2, -1, 0, \) and \( 1 \).
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For \( x = -3 \): \[ y = 3(-3)^2 - 6(-3) + 9 = 3(9) + 18 + 9 = 27 + 18 + 9 = 54 \] (This value is already provided.)
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For \( x = -2 \): \[ y = 3(-2)^2 - 6(-2) + 9 = 3(4) + 12 + 9 = 12 + 12 + 9 = 33 \] (This value is already provided.)
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For \( x = -1 \): \[ y = 3(-1)^2 - 6(-1) + 9 = 3(1) + 6 + 9 = 3 + 6 + 9 = 18 \]
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For \( x = 0 \): \[ y = 3(0)^2 - 6(0) + 9 = 0 + 0 + 9 = 9 \]
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For \( x = 1 \): \[ y = 3(1)^2 - 6(1) + 9 = 3(1) - 6 + 9 = 3 - 6 + 9 = 6 \]
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For \( x = 2 \): \[ y = 3(2)^2 - 6(2) + 9 = 3(4) - 12 + 9 = 12 - 12 + 9 = 9 \] (This value is already provided.)
Now we can complete the table:
| x | y | |----|----| | -3 | 54 | | -2 | 33 | | -1 | 18 | | 0 | 9 | | 1 | 6 | | 2 | 9 |
Next, we find the high or low point of the parabola. Since the coefficient of \( x^2 \) (3) is positive, the parabola opens upwards, meaning it will have a low point (minimum).
The vertex of the parabola given by \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] In our case, \( a = 3 \) and \( b = -6 \): \[ x = -\frac{-6}{2(3)} = \frac{6}{6} = 1 \]
We substitute \( x = 1 \) back into the quadratic equation to find \( y \): \[ y = 3(1)^2 - 6(1) + 9 = 3 - 6 + 9 = 6 \]
Thus, the low point of the parabola is: \[ \text{(1, 6)} \]
Therefore, the final answer is: \[ \text{The low point of this parabola is } (1, 6). \]
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